We investigate the boundedness and the compactness of the
mean operator matrix acting on the weighted Hardy spaces.

1. Introduction

First in the following, we generalize the definitions coming in [1]. Let β={β(n)} be a sequence of positive numbers with β(0)=1 and 1<p<∞. We consider the space of sequences f={f̂(n)}n=0∞ such that‖f‖p=‖f‖βp=∑n=0∞|f̂(n)|pβ(n)p<∞.
The notationf(z)=∑n=0∞f̂(n)zn
will be used whether or not the series converges for any value of z. These are called formal power series and the set of such series is denoted by Hp(β). Let f̂k(n)=δk(n). So fk(z)=zk and then {fk}k is a basis such that ∥fk∥=β(k). Recall that Hp(β) is a reflexive Banach space with norm ∥·∥β and the dual of Hp(β) is Hq(βp/q) where 1/p+1/q=1 and βp/q={β(n)p/q} [2]. For some other sources on this topic see [1–12].

The study of weighted Hardy spaces lies at the interface of analytic function theory and operator theory. As a part of operator theory, research on weighted Hardy spaces is of fairly recent origin, dating back to valuable work of Allen Shields [1] in the mid- 1970s. The mean operator matrix has been the focus of attention for several decades and many of its properties have been studied. Some of basic and useful works in this area are due to Browein et al. [13–16], which are pretty large works that contain a number of interesting results and indeed they are mainly of auxiliary nature. Also, some properties of mean operator matrices have been studied recently by Lashkaripour on weighted sequence spaces [17–20]. In this paper, we have given conditions under which the mean operator matrix is bounded and compact as an operator acting on weighted Hardy spaces. More details of our works are as follows: the idea of Theorem 2.6 comes from [16]. In Theorem 2.9, we extend the method used in [20, Theorem 1.2] to show the boundedness of the mean operator matrix acting on the weighted Hardy spaces. Some inequalities are useful to find a bound for the mean operator matrix acting on weighted Hardy spaces [21–26]. For example the inequality proved in [26, Theorem 8] is used in the proof of Theorem 2.11.

2. Main Results

In this section we define an operator acting on Hp(β) and then we will investigate its boundedness and compactness on Hp(β).

Definition 2.1.

Let {an} be a sequence of positive numbers and define
An=∑i=0naiβ(i)p.
The mean operator matrix associated with the sequence {an} is represented by the matrix A=[ank]n,k and is defined by
ank={akβ(n)pAn,0≤k≤n,0,k>n.

From now on, by A we denote the mean operator matrix associated with the fixed sequence {an} as in Definition 2.1.

Theorem 2.2 (see [<xref ref-type="bibr" rid="B26">12</xref>, Theorem 1]).

If 0<an≤an+1 for all integers n≥0, then A is a bounded operator on Hp(β).

Theorem 2.3 (see [<xref ref-type="bibr" rid="B26">12</xref>, Theorem 2]).

Let 1/p+1/q=1 and bn>0 for n=0,1,… If
M1=supn≥0∑k=0nakβ(n)p+1Anβ(k)(bkbn)1/p<∞,M2=supk≥0∑n=k∞akβ(n)p+1Anβ(k)(bnbk)1/q<∞,
then A=[ank]n,k is a bounded operator on Hp(β) and ∥A∥≤M11/qM21/p.

Recall that if an,bn are two positive sequences, by an~bn, we mean that an/bn→1 whenever n→∞. Also, we write an=o(bn), if an/bn→0 as n→∞.

Corollary 2.4.

Let limn→∞nan/An be finite and 1/p+1/q=1. If
supn≥0∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/p<∞,supk≥0∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/q<∞,
then A is a bounded operator on Hp(β).

Proof.

Put limn→∞nan/An=β. Then nan/β~An and so
∑k=0nakβ(n)p+1Anβ(k)(bkbn)1/p~β∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/pasn⟶∞,∑n=k∞akβ(n)p+1Anβ(k)(bnbk)1/q~β∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/qask⟶∞.
On the other hand
supn≥0∑k=0nakβ(n)p+1nanβ(k)(bkbn)1/p<∞,supk≥0∑n=k∞akβ(n)p+1nanβ(k)(bnbk)1/q<∞,
thus Theorem 2.3 implies that A is a bounded operator on Hp(β).

Lemma 2.5.

Suppose that ncan/β(n) is eventually increasing when the constant c>1-γ, and eventually decreasing when c<1-γ. Let
S1(n)=1n∑k=1nakβ(n)anβ(k)(kn)-1/p,S2(k)=k1/q∑n=k∞akβ(n)anβ(k)1n1/(q+1).
If γ>1/p, then limn→∞S1(n)=limk→∞S2(k)=1/(γ-1/p).

Proof.

Let 1/p+1/q=1 and c2<1-γ<c1<1. Then in either case there is a positive integer N such that
(kn)-c2<akβ(n)anβ(k)<(kn)-c1
for N≤k≤n. Suppose first that γ>1/p, then
limn→∞n1-1/panβ(n)=∞
and hence
limn→∞1n∑k=1N-1akβ(n)anβ(k)(kn)-1/p=0.
Therefore
limn→∞supS1(n)≤limn→∞1n∑k=Nn(kn)-c1-1/p=∫01x-c1-1/pdx.
By calculus integral we get
∫01x-c-1/pdx=11-c-1/p;c≠1q,
and so
limn→∞infS1(n)≥limn→∞1n∑k=Nn(kn)-c2-δ=∫01x-c2-1/pdx=11-c2-1/p.
Letting c1→1-γ from the right and c2→1-γ from the left, we have
limn→∞S1(n)=1γ-1/p.
Also note that
limk→∞supS2(k)≤limk→∞k1/q∑n=k∞(kn)-c11n1/q+1,limk→∞k1/q∑n=k∞(kn)-c11n1/q+1=11/q-c1.
If c1→1-γ, then 1/q-c1→γ-1/p and similarly we get
limk→∞infS2(k)≥limk→∞k1/q-c2∑n=k∞(1n)1-1/q-c2,limk→∞k1/q-c2∑n=k∞(1n)1-1/q-c2=11/q-c2.
If c2→1-γ, then 1/q-c2→γ-1/p. This completes the proof.

Theorem 2.6.

Let limn→∞nanβ(n)p/An=γ, ncanβ(n)p be eventually monotonic for any constant c, and {β(n)} be bounded. Then A is a bounded operator if 1/γ<p.

Proof.

Let δn=nanβ(n)p/An and suppose first that 0≤γ<∞. Then
n(log(An)-log(An-1))=-nlog(1-δnn)⟶γ
as n→∞, and hence
log(An)-log(A1)=-n∑k=2nlog(1-δkk)=ϵnlogn,
where ϵn→γ. Consequently An=A1nϵn. Now suppose that γ=∞, then for n≥2,
log(An)-log(An-1)=-log(1-δnn)≥δnn
since δn→∞. If M>0, then there is N1∈ℕ such that δn≥M+1 for all n≥N1.

Without loss of the generality suppose that there is a positive real number a>0 such that δn>a for n≤N1. Note that
∑k=2N11k=logN1+c+o(1)-1.
If n>N1, then
∑k=2n1k=logn+c+o(1)-1,∑k=N1+1n1k=logn-logN1.
Also,
∑k=2nδk/klogn≥a(∑k=2N11/k)+(M+1)(∑k=N1+1n1/k)logn,∑k=2N11/k+∑k=N1+1n1/klogn≥M1+1(logn-logN1)+a(logN1+c+o(1)-1)logn=M1+1+(a-M1-1)logN1+a(c+o(1)-1)logn,
for large amount of n last equality greater than M1. Hence
logAn≥∑k=2nδkk=γnlogn,
where γn→∞. It follows that, for any real number c, ncAn=nc+γn. Since
nc-1An~1γncanβ(n)p,
thus ncanβ(n)p is eventually increasing for c>1-γ, and eventually decreasing for c<1-γ. But {β(n)}n is bounded, so there are M1,M2>0 such that M1<β(n)<M2, and
ncanβ(n)=ncanβ(n)pβ(n)p+1,ncanβ(n)pβ(n)p+1≥ncanβ(n)pM1p+1.
This implies that ncan/β(n) is eventually increasing for c>1-γ. Similarly ncan/β(n) is eventually decreasing for c<1-γ. Thus
∑k=1nakβ(n)p+1Anβ(k)(kn)-1/p~γn∑k=1nakβ(n)p+1nanβ(k)(kn)-1/p.
By Lemma 2.5γn∑k=1nakβ(n)p+1nanβ(k)(kn)-1/p
is bounded and so
∑k=1nakβ(n)p+1Anβ(k)(kn)-1/p
is bounded. We can see that
∑n=k∞akβ(n)p+1Anβ(k)(kn)1/q
is also bounded. Now by Theorem 2.3, A is a bounded operator and so the proof is complete.

Lemma 2.7.

Let {an},{tn} be nonnegative sequences with t-1=0. Then for all n∈ℕ one has
∑k=0n(tkak)≤{max0≤k≤n(1n-k+1∑j=Knaj)}(∑k=1n(n-k+1)(tk-tk-1)++t0(n+1)).

Proof.

Employing the summation by parts, we get
∑k=0n(tkak)=∑k=0n(∑j=knaj)(tk-tk-1)≤∑k=0n(∑j=knaj1n-k+1)(tk-tk-1)+(n-k+1).
So
∑k=0n(tkak)≤{max0≤k≤n(1n-k+1∑j=knaj)}(∑k=1n(n-k+1)(tk-tk-1)++t0(n+1)),
and at this time the proof is complete.

Theorem 2.8 (see [<xref ref-type="bibr" rid="B6">26</xref>, Theorem 8]).

Let 1/p+1/q=1, {xn} be a positive sequence, then
∑j=0∞max0≤i≤j(1j-i+1∑k=ijxk)p≤qp(∑k=0∞xkp).

Theorem 2.9.

Let {an} be a positive sequence and
M3=supn≥0(∑k=1nn-k+1An(akβ(k)-ak-1β(k-1))+β(n)p+1+(n+1)a0Anβ(0)β(n)p+1)
be finite. Then A is bounded and ∥A∥≤M3q.

Proof.

Let f(z)=∑n=0∞f̂(n)zn∈Hp(β), thus
A(f)(z)=∑n=0∞(∑k=0nakβ(n)pAnf̂(k))zn.
By definition of ∥·∥β, we have
∑n≥0β(n)p|∑k=0nakβ(n)pAnf̂(k)|p≤∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p.
In Lemma 2.7, consider tk=ak/β(k) and aj=|f̂(j)|β(j). Then
∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p≤∑n≥0{max0≤k≤n1n-k+1∑j=kn|f̂(j)|β(j)}p×(∑k=1nn-k+1An(akβ(k)-ak-1β(k-1))+β(n)p+1+(n+1)a0Anβ(0)β(n)p+1)p.
Now, Theorem 2.8 implies that
∑n≥0{max0≤k≤n(1n-k+1∑j=kn|f̂(j)|β(j))}pM3p≤M3pqp∑k=1∞|f̂(k)|pβ(k)p,
and so we get ∥Af∥≤M3q∥f∥β for all f∈Hp(β). Thus A∈B(Hp(β)) and indeed ∥A∥≤M3q. This completes the proof.

Corollary 2.10.

Let 1/p+1/q=1, ak/β(k)≥ak-1/β(k-1) and
M4=supn≥0∑k=0nakβ(n)p+1β(k)An<∞.
Then A is a bounded operator on Hp(β) and ∥A∥≤M4.

Proof.

Note that
∑n≥0(∑k=0nakβ(n)p+1Anβ(k)|f̂(k)|β(k))p≤∑n≥0{max0≤k≤n1n-k+1∑j=kn|f̂(j)|β(j)}p(∑k=0nakβ(n)p+1β(k)An)p.
Theorem 2.8 implies that
∑n≥0{max0≤k≤n(1n-k+1∑j=kn|f̂(j)|β(j))}pM4p≤M4pqp∑k=1∞|f̂(k)|pβ(k)p,
and so by Theorem 2.9 we obtain ∥Af∥≤qM4∥f∥β for all f∈Hp(β). Thus A∈B(Hp(β)) and indeed ∥A∥≤M4q. This completes the proof.

Now, we characterize compactness of subsets of Hp(β) and then we will investigate compactness of the mean operator matrix on Hp(β).

Theorem 2.11.

Let S be a nonempty subset of Hp(β). Then S is relatively compact if and only if the following hold:

there exists M>0, such that for all ∑n=0∞f̂(n)zn∈S, |f̂(i)β(i)|≤M for all i∈ℕ∪{0};

given ϵ>0, there is n0∈ℕ such that ∑n=n0∞|f̂(n)|pβ(n)p<ϵp for all ∑n=0∞f̂(n)zn∈S.

Proof.

Let S be relatively compact, thus there exist g1,…,gk∈Hp(β) such that
S⊆⋃i=1kB(gi,1).
For every f(z)=∑n=0∞f̂(n)zn∈S, there is gi such that f∈B(gi,1). By Minkowski inequality we get
∑n=0∞|f̂(n)|pβ(n)p≤[(∑n=0∞|f̂(n)-ĝi(n)|pβ(n)p)1/p+(∑n=0∞|ĝi(n)|pβ(n)p)1/p]p≤(‖f-gi‖+‖gi‖)p≤(1+‖gi‖)p≤(1+max{‖gi‖:i=1,…,k})p.
Thus for every f∈S and n∈ℕ∪{0}, we get
|f̂(n)β(n)|≤1+max{‖gi‖:i=1,…,k}.
So (i) holds. Now suppose that ϵ is an arbitrary positive number. Since S is relatively compact, thus there exist h1,…,hk∈Hp(β) such that
S⊆⋃i=1kB(hi,ϵ2).
Since hi∈Hp(β), there exists Ni∈ℕ such that
∑n=Ni∞|ĥi(n)|pβ(n)p<ϵp2p
for i=1,…,k. Put
N0=max{Ni:i=1,…,k},
and consider f∈S. Then there exists i∈{1,…,k}, such that f∈B(hi,ϵ/2). Hence we get
∑n=N0∞|f̂(n)|pβ(n)p≤[(∑n=N0∞|f̂(n)-ĥi(n)|pβ(n)p)1/p+(∑n=N0∞|ĥi(n)|pβ(n)p)1/p]p≤(‖f-hi‖+ϵ2)p≤ϵp.
So (ii) holds.

Conversely, assume that ϵ>0 be given and let (i) and (ii) hold. By condition (ii), there exists n0∈ℕ such that
∑n=n0∞|f̂(n)|pβ(n)p<ϵp2,
for all f∈S. Let Mn0 be the closed linear span of the set {1,z,…,zn0-1} in Hp(β). Consider ℂn0 and Mn0 with norms
‖(z1,…,zn0)‖=(∑n=1n0|zn|pβ(n)p)1/p,
for all (zi)i=1n0∈ℂn0, and
‖∑i=0n0-1aizi‖=(∑i=0n0-1|ai|pβ(i)p)1/p
for all ∑i=0n0-1aizi∈Mn0. Define L:Mn0→ℂn0, by
L(∑i=0n0-1aizi)=(a0,…,an0-1).
Clearly, we can see that L is a bounded linear operator. Now, consider the compact subset
{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}
in ℂn0. Then we have
{∑i=0n0-1f̂(i)zi:∑n=0∞f̂(n)zn∈S}⊆L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}.
Since
L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}
is a compact subspace of Mn0, so there exist g1,…,gk∈Mn0 such that
L-1{(zi)i=1n0:∑i=1n0|zi|pβ(i)p≤n0Mp}∈⋃i=1kB(gi,ϵ21/p).
Hence for every
f∈{∑i=0n0-1f̂(i)zi:f(z)=∑n=0∞f̂(n)zn∈S}
there is i∈{1,…,k} satisfying
∑n=0n0-1|f̂(n)-ĝi(n)|pβ(n)p≤ϵp2.
Also, we have
(‖f-gi‖β)p≤∑n=0n0-1|f̂(n)-ĝi(n)|pβ(n)p+∑n=n0∞|f̂(n)|pβ(n)p≤ϵp2+ϵp2≤ϵp.
Thus, S is relatively compact and so the proof is complete.

Theorem 2.12.

Let the mean matrix operator A be bounded on Hp(β), and
limm→∞(∑n=m∞β(n)p2+pAnp)1/p(∑k=0m(akβ(k))q)1/q=0,
where 1/p+1/q=1. Then A is a compact operator on Hp(β).

Proof.

Let BHp(β) be the closed unit ball of Hp(β). Define S=A(BHp(β)) and note that S is a bounded subset of Hp(β). Put rn=|f̂(n)|an, un=β(n)p2+p/Anp, vk=(β(k)/ak)p, and
Em=(∑n=m∞un)1/p(∑k=0mvk1-q)1/q.
Note that limm→∞Em=0. So for every ϵ>0, there exists m0∈ℕ such that Em<ϵ/(qp-1p)1/p for all m≥m0. Note that if
f(z)=∑k=0∞f̂(k)zk∈BHp(β),
then
Af(z)=∑n=0∞(∑k=0nakβ(n)pf̂(k)An)zn∈S.
Since ∥f∥βp≤1, we have
∑n=m∞|Âf(n)|nβ(n)p≤∑n=m∞β(n)p2+pAnp(∑k=0nak|f̂(k)|)p=∑n=m∞un(∑k=0nrk)p≤ϵp∑k=0∞(rk)pvk≤ϵp∑k=0∞|f̂(k)|pβ(k)p≤ϵp.
Thus by Theorem 2.11, S is compact and so the proof is complete.

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